What is the percent by mass of water in FeCl₂.4H₂O?
Let's suppose that we have 1 mol of our compound. So the first step to solve our problem is to find the molar mass of it. To find the molar mass of our compound, we have to look for the atomic mass of each element.
atomic mass of Fe = 55.85 amu
atomic mass of Cl = 35.45 amu
atomic mass of H = 1.01 amu
atomic mass of O = 16.00 amu
Then we will find the molar mass of the compound of FeCl₂.4H₂O. This compound has 1 atom of Fe, 2 atoms of Cl, 8 atoms of H and 4 atoms of O (since it has 4 molecules of water).
molar mass of FeCl₂.4H₂O = 55.85 + 2 * 35.45 + 8 * 1.01 + 4 * 16.00
molar mass of FeCl₂.4H₂O = 198.83 g/mol
We can also find the molar mass of water.
molar mass of H₂O = 2 * 1.01 + 16.00
molar mass of H₂O = 18.02 g/mol
We supposed that we have a sample of 1 mol of FeCl₂.4H₂O. Then the total mass of our sample is 198.83 g.
total mass of the sample = 1 mol * 198.83 g/mol
total mass of the sample = 198.83 g
One molecule of FeCl₂.4H₂O has 4 molecules of H₂O, so 1 mol of FeCl₂.4H₂O has 4 moles of H₂O. So our sample has 4 moles of H₂O (since we supposed that the sample is 1 mol of FeCl₂.4H₂O). So the mass of water in our sample is:
mass of H₂O = 4 moles of H₂O * 18.02 g/mol
mass of H₂O = 72.08 g
The mass of the sample is 198.83 g, and in that sample there are 72.08 g of H₂O. Using those values we can find the percent by mass of water:
% by mass of H₂O = mass of H₂O/mass of sample * 100
% by mass of H₂O = 72.08 g / (198.83 g) * 100
% by mass of H₂O = 36.25 %
Answer: The percent by mass of water in FeCl₂.4H₂O is 36.25 %